Let $f$ a retraction from $\overline{B}^m$ to $S^{m-1}$of class $C^1$ .For all $t\in(0,1)$ we put: $$\bigtriangledown f_t=(1-t)I+t\bigtriangledown f$$ And : $$v(t)=\int_{B^m}det \bigtriangledown f_t(x)dx $$
why the degree of th polynom $V$ is smaller than $m$ in the $t$ ?
Thank you .
first, note that $\nabla f_t$ is affine linear in $t$.
It is also known that the determinant of a matrix of size $m\times m$ is a polynomial of the components with degree $m$.
So $\det\nabla f_t$ is a polynomial in $t$ with degree $\leq m$ (that also depends on $x$).
integrating over a domain does not change this, since $x$ is the integration variable.